It is known that given primitive (co-prime) integer solutions to,
$$x_1^4+x_2^4+x_3^4+x_4^4 = z^4$$
then there is one $x_i$ such that $z^4-x_i^4$ is divisible by $d_4=5^4$. Additionally, Ward showed that if one of the $x_i$ is zero, then there is the further constraint that $z\pm x_j$ is divisible by $w_4=2^{10}$. For example, we have,
$$673865^4+ 1390400^4+ 2767624^4 = 2813001^4$$
and,
$$z+x_3 = 2813001 + 2767624 = 5^4\cdot 8929$$
$$z-x_1 = 2813001 - 673865= 2^{10}\cdot 2089$$
Theorem: In general, for $k$ $k$th powers and co-prime terms,
$$x_1^k+x_2^k+x_3^k+\dots+x_k^k = z^k\tag1$$
if $k+1$ is prime, then there is one $x_i$ such that $z^k-x_i^k$ is divisible by $d_k = (k+1)^k$.
For $d_4 = 5^4 = 625$, this implies the smallest solution will have an addend $x_i >d_4/2 \approx 312$. In fact, we have,
$$30^4+120^4+\color{blue}{272}^4+315^4 = 353^4$$
and $z+x_3 =353+\color{blue}{272}=5^4$.
Incidentally, this has an analogue for $6$th powers when using seven addends,
$$1344^6+ 23268^6+ 25263^6+ 39088^6+ 48090^6+ \color{blue}{54138}^6+ 54018^6 = 63631^6$$
and $z+x_7 =63631+\color{blue}{54018}=7^6$.
Question: Given $(1)$ where one of the $x_i$ is zero, is there an analogue to Ward's constraint $w_k$ for $k=6$? If there is, what is it for general $k$?
P.S. For odd powers, see also the post for $x_1^5+x_2^5+x_3^5+x_4^5+x_5^5=0$
